# How many points does y=-2x^2+x-3 have in common with the vertex and where is the vertex in relation to the x axis?

Jun 8, 2017

${x}_{\text{vertex}} = + \frac{1}{4}$

The number of points in common with the vertex and the graph is 1

#### Explanation:

This is a quadratic equation and the coefficient of ${x}^{2}$ is negative. Consequently the graph is of form $\cap$ thus the vertex is a maximum

$\textcolor{red}{\text{If}}$ the coefficient has been positive then you would have the general form of $\cup$

The vertex has only one point.

$\textcolor{g r e e n}{\text{A sort of cheat to determine x-vertex}}$
Not really a cheat as it is part of the process for completing the square.

Write as: $\textcolor{\mathrm{da} r k c y a n}{- 2} \left({x}^{2} + \textcolor{m a \ge n t a}{\frac{1}{- 2}} x\right) - 3$

Note that $\left(\textcolor{\mathrm{da} r k c y a n}{- 2}\right) \times \left(\textcolor{m a \ge n t a}{- \frac{1}{2}}\right) x = + x$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \textcolor{m a \ge n t a}{- \frac{1}{2}} = + \frac{1}{4}$